Stable Marriage Problem What criteria to use? Stability. - PowerPoint PPT Presentation

Stable Marriage Problem What criteria to use? Stability. Introduced by Gale and Shapley in a 1962 paper in the American Mathematical Monthly. Consider the couples: Proved useful in many settings, led eventually to 2012 Nobel ◮ Alice and Bob ◮ Maximize number of first choices. Prize in Economics (to Shapley and Roth). ◮ Mary and John ◮ Minimize difference between preference ranks. Original Problem Setting: ◮ Look for stable matchings Bob prefers Mary to Alice. ◮ Small town with n men and n women. Mary prefers Bob to John. ◮ Each woman has a ranked preference list of men. Uh...oh! Unstable pairing. ◮ Each man has a ranked preference list of women. How should they be matched? So.. A stable pairing?? The Stable Marriage Algorithm. Given a set of preferences. Each Day: Is there a stable pairing? 1. Each man proposes to his favorite woman on his list. How does one find it? Produce a pairing where there is no running off! 2. Each woman rejects all but her favorite proposer Consider a variant of this problem: stable roommates. Definition: A pairing is disjoint set of n man-woman pairs. (whom she puts on a string.) A B C D Example: A pairing S = { ( Bob , Alice );( John , Mary ) } . B C A D 3. Rejected man crosses rejecting woman off his list. C A B D Definition: A rogue couple b , g for a pairing S : Stop when each woman gets exactly one proposal. D A B C b and g prefer each other to their partners in S Does this terminate? C D Example: Bob and Mary are a rogue couple in S . ...produce a pairing? ....a stable pairing? Do men or women do “better”? A B

Example. Termination. It gets better every day for women.. Improvement Lemma: If man b proposes to a woman on day k , every future day, she has on a string a man b ′ she likes at least as much as b . Men Women (that is, her options get better) X A 1 2 3 1 C A B Proof: X X B 1 2 3 2 A B C Every non-terminated day a man crossed an item off the list. Ind. Hyp.: P ( j ) ( j ≥ k ) — “Woman has as good an option on X C 2 1 3 3 A C B Total size of lists? n men, n length list. n 2 day j as on day k .” Terminates in at most n 2 + 1 steps! Base Case: P ( k ) : either she has no one/worse on a string (so Day 1 Day 2 Day 3 Day 4 Day 5 X X , C puts b or better on a string), or she has someone better already. 1 A, B A A C C X X Assume P ( j ) . Let ˆ b be man on string on day j ≥ k . So ˆ 2 C B, C B A,B A b is as 3 B good as b . On day j + 1, man ˆ b will come back (and possibly others). Woman can choose ˆ b just as well, or pick a better option. = ⇒ P ( j + 1 ) . Pairing when done. Pairing is Stable. Good for men? women? Lemma: There is no rogue couple for the pairing formed by Is the SMA better for men? for women? Lemma: Every man is matched at end. stable marriage algorithm. Definition: A pairing is x -optimal if x ′ s partner Proof: Proof: is its best partner in any stable pairing. If not, a man b must have been rejected n times. Assume there is a rogue couple; ( b , g ∗ ) Definition: A pairing is x -pessimal if x ′ s partner Every woman has been proposed to by b , is its worst partner in any stable pairing. b likes g ∗ more than g . g ∗ b ∗ and Improvement lemma Definition: A pairing is man optimal if it is x -optimal for all men x . = ⇒ each woman has a man on a string. ..and so on for man pessimal, woman optimal, woman pessimal. g ∗ likes b more than b ∗ . g b and each man on at most one string. Claim: The optimal partner for a man must be first in his preference list. n women and n men. Same number of each. Man b proposes to g ∗ before proposing to g . True? False? False! = ⇒ b must be on some woman’s string! So g ∗ rejected b (since he moved on) Subtlety here: Best partner in any stable pairing. Contradiction. As well as you can in a globally stable solution! By improvement lemma, g ∗ likes b ∗ better than b . Question: Is there a even man or woman optimal pairing? Contradiction!

SMA is optimal! How about for women? Residency Matching.. For men? For women? Theorem: SMA produces a man-optimal pairing. Theorem: SMA produces woman-pessimal pairing. Proof: Assume not: there are men who do not get their optimal woman. T – pairing produced by SMA. Let t be first day any man b gets rejected S – worse stable pairing for woman g . by his optimal woman g who he is paired with The method was used to match residents to hospitals. In T , ( g , b ) is pair. in some stable pairing S . Hospital optimal.... In S , ( g , b ∗ ) is pair. b is paired with someone else, say g ∗ . Let g put b ∗ on a string in place of b on day t = ⇒ g prefers b ∗ to b ..until 1990’s...Resident optimal. g likes b ∗ less than she likes b . By choice of day t , b ∗ has not yet been rejected by his optimal Variations: couples! T is man optimal, so b likes g more than g ∗ , his partner in S . woman. Therefore, b ∗ prefers g to optimal woman, and hence to his partner ( g , b ) is Rogue couple for S g ∗ in S . S is not stable. Rogue couple for S . Contradiction. So S is not a stable pairing. Contradiction. Recap: S - stable. ( b ∗ , g ∗ ) ∈ S . But ( b ∗ , g ) is rogue couple! Used Well-Ordering principle... Fun stuff from the Fall 2014 offering... Follow the link. Link